Question

Let N denote the set of all natural numbers. Define two binary relations on N as $R_1=\left\{\right(x,y)ϵN\times N:2x+y=10\}$ and $R_2=\left\{\right(x,y)ϵN\times N:x+2y=10\}$. Then.

• A. Both $R_1$ and $R_2$ are transitive relations
• B. Both $R_1$ and $R_2$ are symmetric relations
• C. Range of $R_2$ is $\{1,2,3,4\}$
• D. Range of $R_1$ is $\{2,4,8\}$

Answer 1

The correct option is C Range of $R_2$ is $\{1,2,3,4\}$

Define two binary relations on N as $R_1=\left\{\right(x,y)ϵN\times N:2x+y=10\}$ and $R_2=\left\{\right(x,y)ϵN\times N:x+2y=10\}$

From $R_1$, $2x+y=10$ and $x,y\in N$

So, possible values for $x$ and $y$ are:

$x=1,y=8$ i.e $(1,8)$

$x=2,y=6$ i.e $(2,6)$

$x=3,y=4$ i.e $(3,4)$

$x=4,y=2$ i.e $(4,2)$

$R_1=\left\{\right(1,8),(2,6),(3,4),(4,2\left)\right\}$

Therefore, Range of $R_1$ is $\{2,4,6,8\}$

$R_1$ is not symmetric

Also, $R_1$ is not transitive because $(3,4),(4,2)\in R_1$ but $(3,2)\notin R_1$

Thus, options A,B and D are incorrect.

From $R_2$, $x+2y=10$ and $x,y\in N$

So, possible values for $x$ and $y$ are:

$x=8,y=1$ i.e $(8,1)$

$x=6,y=2$ i.e $(6,2)$

$x=4,y=3$ i.e $(4,3)$

$x=2,y=4$ i.e $(2,4)$

$R_2=\left\{\right(8,1),(6,2),(4,3),(2,4\left)\right\}$

Therefore, Range of $R_2$ is $\{1,2,3,4\}$

$R_2$ is not symmetric

Hence, option C is correct.